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Chapter 6: Problem 34

Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of \(1 .\) In each case, draw a graph, then find the probability of the given bone density test scores. If using technology instead of Table A-2, round answers to four decimal places. $$ \text { Greater than }-3.75 $$

Short Answer

Expert verified
The probability is 0.9999.

Step by step solution

01

- Identify Given Information

The mean is \(\bar{x} = 0\) and the standard deviation is \( \sigma = 1\). The score we need to find the probability for is greater than \(\text{-3.75}\).
02

- Standardize the Score

Since the test scores are normally distributed and we are given a mean and standard deviation, we convert the score to a z-score. Since the mean is \(\bar{x} = 0\) and standard deviation is \( \sigma = 1\), the z-score formula simplifies to \( z = \frac{x - \bar{x}}{\sigma} = \frac{-3.75 - 0}{1} = -3.75\).
03

- Use the Z-Table or Technology

Using a Z-table or technology (such as a statistical calculator or software), find the probability corresponding to the z-score \( z = -3.75\). This value is the area to the left of \( z = -3.75\).
04

- Calculate the Probability

The table or software shows the area to the left of \( z = -3.75\) is approximately 0.000088. Since we need the area to the right, we subtract this value from 1: \( P(Z > -3.75) = 1 - P(Z < -3.75) = 1 - 0.000088 = 0.9999\).
05

- Draw the Graph

Sketch the standard normal distribution curve, place -3.75 on the horizontal axis, and shade the area to the right. This shaded region represents the probability we found.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

normal distribution
Understanding the normal distribution is crucial for solving probability problems like bone density tests. A normal distribution is a continuous probability distribution shaped like a symmetrical bell curve. The mean (average) is at the center, and it splits the curve into two equal halves. Most values cluster around the mean, with fewer values as you move towards the tails. The standard deviation determines how spread out the values are around the mean. In this exercise, the mean (渭) is 0, and the standard deviation (蟽) is 1. This is known as a standard normal distribution.
z-score calculation
The z-score tells us how many standard deviations an element is from the mean. It standardizes different data points, making them comparable. The formula to calculate the z-score is \[ z = \frac{x - \bar{x}}{\sigma} \]where x is the data point, \bar{x} is the mean, and \sigma\ is the standard deviation. In our bone density test score problem, the score -3.75 is converted using the formula \[ z = \frac{-3.75 - 0}{1} = -3.75 \]This tells us that -3.75 is 3.75 standard deviations below the mean.
standard normal distribution curve
The standard normal distribution curve is a bell-shaped graph with a mean of 0 and a standard deviation of 1. All standard normal curves are identical aside from scaling. To visualize the problem, graph the curve using the x-axis to represent z-scores. The z-score of -3.75 falls far to the left. In this case, we shade the area to the right of -3.75, representing the probability we are interested in. The total area under the curve equals 1, meaning the probabilities add up to 100%.
probability calculation
Once the z-score is calculated, we use it to find the relevant probability. This is often done using a Z-table or technology like computational software. The Z-table gives the probability for areas to the left of a z-score \[ P(Z < -3.75) \]For z = -3.75, the left-side area is approximately 0.000088. Since we want the area to the right \[ P(Z > -3.75) \] We subtract this value from 1: \[ P(Z > -3.75) = 1 - P(Z < -3.75) = 1 - 0.000088 = 0.9999 \]This means there is a 99.99% chance that a randomly selected test score will be greater than -3.75. Remember to visualize this on the curve by shading the appropriate area.

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Most popular questions from this chapter

Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of \(1 .\) In each case, draw a graph, then find the probability of the given bone density test scores. If using technology instead of Table A-2, round answers to four decimal places. $$ \text { Greater than }-3.05 $$

There are about 4200 college presidents in the United States, and they have annual incomes with a distribution that is skewed instead of being normal. Many different samples of 40 college presidents are randomly selected, and the mean annual income is computed for each sample. a. What is the approximate shape of the distribution of the sample means (uniform, normal, skewed, other)? b. What value do the sample means target? That is, what is the mean of all such sample means?

Use the data in the table below for sitting adult males and females (based on anthropometric survey data from Gordon, Churchill, et al.). These data are used often in the design of different seats, including aircraft seats, train seats, theater seats, and classroom seats. (Hint: Draw a graph in each case.) $$ \begin{aligned} &\text { Sitting Back-to-Knee Length (inches) }\\\ &\begin{array}{l|c|c|c} \hline & \text { Mean } & \text { St. Dev. } & \text { Distribution } \\ \hline \text { Males } & 23.5 \text { in. } & 1.1 \text { in. } & \text { Normal } \\ \hline \text { Females } & 22.7 \text { in. } & 1.0 \text { in. } & \text { Normal } \\ \hline \end{array} \end{aligned} $$ For females, find the first quartile \(Q_{1}\), which is the length separating the bottom \(25 \%\) from the top \(75 \%\).

Mensa requires a score in the top \(2 \%\) on a standard intelligence test. The Wechsler IQ test is designed for a mean of 100 and a standard deviation of 15, and scores are normally distributed. a. Find the minimum Wechsler IQ test score that satisfies the Mensa requirement. b. If 4 randomly selected adults take the Wechsler IQ test, find the probability that their mean score is at least 131 . c. If 4 subjects take the Wechsler test and they have a mean of 132 but the individual scores are lost, can we conclude that all 4 of them are eligible for Mensa?

The following are the values of net worth (in thousands of dollars) of recent members of the executive branch of the U.S. government. Test these values for normality, then take the logarithm of each value and test for normality. What do you conclude? \(\begin{array}{lllllllllll}237,592 & 16,068 & 15,350 & 11,712 & 7304 & 6037 & 4483 & 4367 & 2658 & 1361 & 311\end{array}\)
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